3.549 \(\int \frac{(a+b x^2)^{5/2} (A+B x^2)}{x^6} \, dx\)

Optimal. Leaf size=152 \[ \frac{b^2 x \sqrt{a+b x^2} (5 a B+2 A b)}{2 a}+\frac{1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac{b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(3/2))/(3*a*x) - ((2*A*b + 5*a*
B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/2

________________________________________________________________________________________

Rubi [A]  time = 0.0624304, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {453, 277, 195, 217, 206} \[ \frac{b^2 x \sqrt{a+b x^2} (5 a B+2 A b)}{2 a}+\frac{1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac{b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(3/2))/(3*a*x) - ((2*A*b + 5*a*
B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/2

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx &=-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}-\frac{(-2 A b-5 a B) \int \frac{\left (a+b x^2\right )^{5/2}}{x^4} \, dx}{5 a}\\ &=-\frac{(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac{(b (2 A b+5 a B)) \int \frac{\left (a+b x^2\right )^{3/2}}{x^2} \, dx}{3 a}\\ &=-\frac{b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac{(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac{\left (b^2 (2 A b+5 a B)\right ) \int \sqrt{a+b x^2} \, dx}{a}\\ &=\frac{b^2 (2 A b+5 a B) x \sqrt{a+b x^2}}{2 a}-\frac{b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac{(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac{1}{2} \left (b^2 (2 A b+5 a B)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{b^2 (2 A b+5 a B) x \sqrt{a+b x^2}}{2 a}-\frac{b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac{(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac{1}{2} \left (b^2 (2 A b+5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{b^2 (2 A b+5 a B) x \sqrt{a+b x^2}}{2 a}-\frac{b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac{(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac{1}{2} b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0364749, size = 84, normalized size = 0.55 \[ \frac{a \sqrt{a+b x^2} (-5 a B-2 A b) \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^2}{a}\right )}{15 x^3 \sqrt{\frac{b x^2}{a}+1}}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

-(A*(a + b*x^2)^(7/2))/(5*a*x^5) + (a*(-2*A*b - 5*a*B)*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((
b*x^2)/a)])/(15*x^3*Sqrt[1 + (b*x^2)/a])

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 251, normalized size = 1.7 \begin{align*} -{\frac{B}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Bb}{3\,{a}^{2}x} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{b}^{2}Bx}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{2}Bx}{3\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}Bx}{2}\sqrt{b{x}^{2}+a}}+{\frac{5\,Ba}{2}{b}^{{\frac{3}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }-{\frac{A}{5\,a{x}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{2\,Ab}{15\,{a}^{2}{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{8\,A{b}^{2}}{15\,{a}^{3}x} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{8\,A{b}^{3}x}{15\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,A{b}^{3}x}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{A{b}^{3}x}{a}\sqrt{b{x}^{2}+a}}+A{b}^{{\frac{5}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x)

[Out]

-1/3*B/a/x^3*(b*x^2+a)^(7/2)-4/3*B*b/a^2/x*(b*x^2+a)^(7/2)+4/3*B*b^2/a^2*x*(b*x^2+a)^(5/2)+5/3*B*b^2/a*x*(b*x^
2+a)^(3/2)+5/2*B*b^2*x*(b*x^2+a)^(1/2)+5/2*B*b^(3/2)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-1/5*A*(b*x^2+a)^(7/2)/a/x
^5-2/15*A*b/a^2/x^3*(b*x^2+a)^(7/2)-8/15*A*b^2/a^3/x*(b*x^2+a)^(7/2)+8/15*A*b^3/a^3*x*(b*x^2+a)^(5/2)+2/3*A*b^
3/a^2*x*(b*x^2+a)^(3/2)+A*b^3/a*x*(b*x^2+a)^(1/2)+A*b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.80469, size = 525, normalized size = 3.45 \begin{align*} \left [\frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{b} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (15 \, B b^{2} x^{6} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{60 \, x^{5}}, -\frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{-b} x^{5} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (15 \, B b^{2} x^{6} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="fricas")

[Out]

[1/60*(15*(5*B*a*b + 2*A*b^2)*sqrt(b)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(15*B*b^2*x^6 -
2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5, -1/30*(15*(5*B*a*b +
2*A*b^2)*sqrt(-b)*x^5*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (15*B*b^2*x^6 - 2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*
a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5]

________________________________________________________________________________________

Sympy [B]  time = 7.30971, size = 292, normalized size = 1.92 \begin{align*} - \frac{A \sqrt{a} b^{2}}{x \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{5 x^{4}} - \frac{11 A a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{15 x^{2}} - \frac{8 A b^{\frac{5}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{15} + A b^{\frac{5}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )} - \frac{A b^{3} x}{\sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{2 B a^{\frac{3}{2}} b}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{B \sqrt{a} b^{2} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} - \frac{2 B \sqrt{a} b^{2} x}{\sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{3 x^{2}} - \frac{B a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3} + \frac{5 B a b^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**6,x)

[Out]

-A*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 11*A*a*b**(3/2)*sqrt(a
/(b*x**2) + 1)/(15*x**2) - 8*A*b**(5/2)*sqrt(a/(b*x**2) + 1)/15 + A*b**(5/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**3
*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - 2*B*a**(3/2)*b/(x*sqrt(1 + b*x**2/a)) + B*sqrt(a)*b**2*x*sqrt(1 + b*x**2/a)/
2 - 2*B*sqrt(a)*b**2*x/sqrt(1 + b*x**2/a) - B*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - B*a*b**(3/2)*sqrt(a
/(b*x**2) + 1)/3 + 5*B*a*b**(3/2)*asinh(sqrt(b)*x/sqrt(a))/2

________________________________________________________________________________________

Giac [B]  time = 1.16989, size = 433, normalized size = 2.85 \begin{align*} \frac{1}{2} \, \sqrt{b x^{2} + a} B b^{2} x - \frac{1}{4} \,{\left (5 \, B a b^{\frac{3}{2}} + 2 \, A b^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \,{\left (45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} B a^{2} b^{\frac{3}{2}} + 45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} A a b^{\frac{5}{2}} - 150 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} B a^{3} b^{\frac{3}{2}} - 90 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} A a^{2} b^{\frac{5}{2}} + 200 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} B a^{4} b^{\frac{3}{2}} + 140 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} A a^{3} b^{\frac{5}{2}} - 130 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{5} b^{\frac{3}{2}} - 70 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} A a^{4} b^{\frac{5}{2}} + 35 \, B a^{6} b^{\frac{3}{2}} + 23 \, A a^{5} b^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b^2*x - 1/4*(5*B*a*b^(3/2) + 2*A*b^(5/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(4
5*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(3/2) + 45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a*b^(5/2) - 150*(sqrt(b
)*x - sqrt(b*x^2 + a))^6*B*a^3*b^(3/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(5/2) + 200*(sqrt(b)*x - s
qrt(b*x^2 + a))^4*B*a^4*b^(3/2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^3*b^(5/2) - 130*(sqrt(b)*x - sqrt(b*
x^2 + a))^2*B*a^5*b^(3/2) - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^4*b^(5/2) + 35*B*a^6*b^(3/2) + 23*A*a^5*b^(
5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5